Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.awt.*;
import java.io.PrintWriter;
import java.util.*;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt();
        double coins[]=new double[n+1];
        for (int i=1;i<=n;i++)coins[i]=sc.nextDouble();
        double dp[][]=new double[n+1][n+1];
        dp[0][0]=1.0;
        for (int i=1;i<=n;i++){
            for (int j=0;j<=i;j++){
                if (j==0)dp[i][j]=dp[i-1][j]*(1.0-coins[i]);
                else {
                    dp[i][j]=dp[i-1][j]*(1.0-coins[i])+dp[i-1][j-1]*coins[i];
                }
            }
        }
        double ans=0.0;
        for (int i=(n+1)/2;i<=n;i++){
            ans+=dp[n][i];
        }
        System.out.println(ans);
    }
}